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mr_godi [17]
3 years ago
14

A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho

w long did it take the ball to fall to the ground?
Physics
1 answer:
il63 [147K]3 years ago
6 0
Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second


Hope it helps...!!!
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y=\frac{1}{2} gt^2

rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.

t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2}  \\ =10.096 s

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